Tricky Trig

Geometry Level 4

The sum of all positive solutions of 2 cos ⁡ 2 x ( cos ⁡ 2 x − cos ⁡ 2014 π 2 x ) = cos ⁡ 4 x − 1 2\cos 2x \left( \cos 2x- \cos\frac { 2014 \pi ^ 2 }{ x } \right) =\cos { 4x } -1 is k π k\pi . Find k k .


The answer is 1080.

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Shriya Mandarapu by shriya mandarapu , 24, India

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1 solution

Shriya Mandarapu
May 18, 2014

Rewrite

cos ⁡ 4 x − 1 \cos{4x} - 1 as 2 cos ⁡ 2 2 x − 2. 2\cos^2{2x} - 2.

Now let a = cos ⁡ 2 x \cos{2x} , and let b = cos ⁡ ( 2014 π 2 x ) \cos{\left( \frac{2014\pi^2}{x} \right) }

. We have 2a(a - b) = 2 a 2 { a }^{ 2 } - 2 and ab = 1

Notice that either a = 1 and b = 1 or a = -1 and b = -1.

For the first case, a = 1 only when x = k π k\pi and k is an integer. b = 1 when 2014 π 2 k π \frac{2014\pi^2}{k\pi} is an even multiple of π \pi , and since 2014 = 2 ∗ 19 ∗ 53 2\ast 19\ast 53 , b =1 only when k is an odd divisor of 2014.

This gives us these possible values for x: x= π \pi , 19 π 19\pi , 53 π 53\pi , 1007 π 1007\pi

For the case where a = -1, cos ⁡ 2 x \cos{2x} = -1, so x = m π 2 \frac{m\pi}{2} , where m is odd. 2014 π 2 m π 2 \frac{2014\pi^2}{\frac{m\pi}{2}} must also be an odd multiple of π \pi in order for b to equal -1, so 4028 m \frac{4028}{m} must be odd.

We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for m, and therefore no cases where a = -1 and b = -1. Therefore, the sum of all our possible values for x is π + 19 π + 53 π + 1007 π = 1080 π \pi +19\pi +53\pi +1007\pi =1080\pi

8 Helpful 0 Interesting 0 Brilliant 1 Confused

Most difficult part is factorising 1007........ I thought it would be prime how to check..

Dhruv Joshi - 4 years, 2 months ago

I've edited the Latex in your question. Can you review it for accuracy?

Calvin Lin Staff - 7 years ago

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yeah sure..i don't think there are any errors..

shriya mandarapu - 7 years ago

Nice problem !

Dinesh Chavan - 7 years ago

This is 2014 AMC 12B, #25.

Michael Tang - 7 years ago

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Yep that is right

Mardokay Mosazghi - 7 years ago

Distributing 2 C o s 2 x 2Cos2x , and using trig identity we get 2 ∗ C o s 2 2 x − 2 ∗ C o s 2 x ∗ C o s 2014 π 2 x = 2 C o s 2 2 x − 2. G i v e s C o s 2 x ∗ C o s 2014 π 2 x = 1 F o r a l l i n t e g e r v a l u e o f x , C o s 2 x = 1. S o C o s 2014 π 2 x m u s t = 1. L e t m b e a + t i v e i n t e g e r . x ≠ 0. 2014 π 2 m π = 2014 π m = 2 π m ∗ 1007 S o C o s 2014 π 2 x = 1 i f m ∣ 1007 ⟹ m = 1 , 19 , 53 , 1007. k = 1 + 19 + 53 + 1007 = 1080 2*Cos^22x -2*Cos2x * Cos\frac{2014\pi^2}{x}=2Cos^22x - 2.~ \\Gives~Cos2x * Cos\frac{2014\pi^2}{x} =1\\For~all~ integer ~ value~ of~x,~~~~Cos2x=1.~~~So~~Cos\frac{2014\pi^2}{x}~must~=1.\\Let~m~be~a~+tive~integer. \\x\neq 0.~~ \frac{2014\pi^2}{m\pi} =\frac{2014\pi}{m} =\frac{2\pi}{m}*1007~~So~Cos\frac{2014\pi^2}{x} =1~if~ \\ m|1007~~\implies~m=1, 19, 53, 1007.~~~~k=1+19+53+1007=\boxed{1080}

Niranjan Khanderia - 6 years, 6 months ago

Made a silly mistake! but note that by expanding both sides we get cos2014π^2/x = cos 0 => x = 1007/n so 19π, 53π and 1007π are the positive solutions

Advaith Kumar - 1 year ago

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